Question

The freezing point of benzene decreases by ${0.45}^{o}C$ when $0.2g$ of acetic acid is added to $20g$ of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be :
($K_f$ for benzene $=5.12Kkgmo{l}^{-1}$)

• A. $80.4\%$
• B. $74.6\%$
• C. $94.6\%$
• D. $64.6\%$

Answer 1

Answer: (C)

$94.6\%$

Molality $m=1000g/kg\times \frac{0.2g}{(60g/mol\times 20g)}=0.1667m$
$\Delta Tf=i{K}_{f}m$
$0.45=i\times 5.12\times 0.1667$
The vant Hoff factor $i=0.527$
The degree of association $\alpha =\frac{(1-i)}{(1-1/n)}$
The degree of association $\alpha =\frac{(1-0.527)}{(1-1/2)}=0.946$
Percent association $=0.946\times 100=94.6$
Hence, the option (C) is the correct answer.